HSF.BF.B.3 math practice. Learn by doing.

Adding $7$ outside the function increases every output by $7$, so the graph shifts $7$ units up.

To shift left, $x$ gets smaller but the height stays the same, so $4$ is added inside $f$. That is why you see $(x + 4)$.

An upward shift of $6$ replaces $y$ with $y - 6$. At the $y$-intercept, $x = 0$, so $3(y - 6) = 9$. Then $3y - 18 = 9$, $3y = 27$, and $y = 9$.

Subtracting $3$ outside the function decreases every output by $3$, so the graph shifts $3$ units down.

Multiplying the outputs by $2$ doubles every $y$-value and leaves the $x$-values unchanged. The graph is stretched vertically by a factor of $2$.

A right shift of $5$ replaces $x$ with $x - 5$. At the $x$-intercept, $y = 0$, so $6(x - 5) = 18$. Then $6x - 30 = 18$, $6x = 48$, and $x = 8$.

Replacing $x$ with $x + 5$ is equivalent to $f(x - (-5))$, so the input must be $5$ smaller to produce the same output and the graph shifts $5$ units left.

Subtracting $5$ outside $f$ decreases every output by $5\textsf{.}$ The graph shifts down $5$ units.